\(M=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-xy-y^2+2y+y+x-2+2019\)

\(\Rightarrow M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(y+x-2\right)+2019\)

\(\Rightarrow M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)

\(\Rightarrow M=\left(x^2-y+1\right).0+2019\)

\(\Rightarrow M=0+2019\)

\(\Rightarrow M=2019\)

M = x³ + x²y - 2x² - xy - y² + 3y + x + 2017

M = (x³ + x²y - 2x²) - (xy + y² - 2y) + (x + y - 2) + 2019

M = x². (x + y - 2) - y(x + y - 2) + (x + y - 2) + 2019 = 2019

\(M = x^3 + x^2y - 2x^2 - xy - y^2 + 3y + x + 2017.\)

\(M=(x^3+x^2y-2x^2)-(xy-y^2+2y)+(x+y-2)+2019\)

\(M=x^2.(x+y-2)-y.(x-y+2)+(x+y-2)+2019\)

\(M=x^2.0-y.0+0+2019\)

\(M=0-0+0+2019\)

\(M=2019\)

$M={}^{}+{}^{}y-2{}^{}-xy-{}^{}+3y+x+2017.$

$M=({}^{}+{}^{}y-2{}^{})-(xy-{}^{}+2y)+(x+y-2)+2019$

$M={}^{}.(x+y-2)-y.(x-y+2)+(x+y-2)+2019$

$M={}^{}.0-y.0+0+2019$

Bài 1:

Ta thấy: $(x+\frac{1}{2})^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow (x+\frac{1}{2})^2+\frac{5}{4}\geq \frac{5}{4}$

Vậy gtnn của biểu thức là $\frac{5}{4}$

Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}$

Bài 2:

$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)

\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)

ko trả lời thì thôi đừng nhắn bậy

đúng ko trả lời cứ nhắn bậy

M=(x3+x2y-2x2)+(2y-y2-xy)+(x+y-2)+2020

M=x2(x+y-2)+y(2-y-x)+(x+y-2)+2020

M=x2.0+y.0+0+2020

M=2020

Vậy M=2020

không hiểu chỗ nào hỏi mình nha!

Bài làm

Ta có: P = x³ + x²y - 2x² - xy - y² + 3y + x + 2017

          P = x³ + x²y - 2x² - xy - y² + 2y + y + x + 2017

          P = ( x³ + x²y − 2x² ) − ( xy + y² − 2y ) + ( x + y − 2 ) + 2019

          P = x²( x + y − 2 ) − y( x + y − 2 ) + ( x + y − 2 ) + 2019

Mà x + y = 2 => x + y - 2 = 0

Thay x + y - 2 = 0 và đa thức P, ta được:

P = x² . 0 - y . 0 + 0 + 2019

P = 0 - 0 + 0 + 2019

P = 2019

Vậy P = 2019 tại x + y = 2

# Học tốt #

\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(P=\left(x^3+x^2y-2x^2\right)+\left(-xy-y^2+2y\right)+\left(x+y-2\right)+2019\)

\(P=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+2019\)

\(P=\left(x^2-y+1\right)\left(x+y-2\right)+2019\)

\(P=0+2019=2019\)

Ta có

\(P=x^3+x^2y-2x^2-xy-y^2+3y+x+2017\)

\(\Leftrightarrow x^3+x^2y-2x^2-xy-y^2+2y+y+x+2017\)

\(\Leftrightarrow\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+2019\)

\(\Leftrightarrow x^2\cdot\left(x+y-2\right)-y\cdot\left(x+y-2\right)+\left(x+y-2\right)+2019\)

Ta có \(x+y=2\Rightarrow x+y-2=0\)

\(\Rightarrow P=2019\)